# 103. Binary Tree Zigzag Level Order Traversal

Given a binary tree, return thezigzag level ordertraversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree`[3,9,20,null,null,15,7]`,

``````    3
/ \
9  20
/  \
15   7
``````

return its zigzag level order traversal as:

``````[
[3],
[20,9],
[15,7]
]
``````

Thoughts:

1. order to expanding which node's child in the current level -> using stack<TreeNode*> to reverse the order.
2. for each Node in the current level, decide weather to first push left or right into the queue:
1. curDepth is odd -> next level even -> Normal Order
2. curDepth is even -> next level odd -> reverse Order

Code

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
queue <TreeNode *> q;
stack<TreeNode *> levelNode;

q.push(root);
int curDepth = 0;
while(!q.empty()){
int len = q.size();
vector<int> level;
for(int i = 0 ; i < len; i ++){
TreeNode * cur = q.front();
// visit
level.push_back(cur->val);
levelNode.push(cur);
q.pop();
}
for(int i = 0; i < len; i++){
TreeNode * cur = levelNode.top();levelNode.pop();
if(curDepth %2 ==0){
if(cur->right) q.push(cur->right);
if(cur->left) q.push(cur->left);
}
else{

if(cur->left) q.push(cur->left);
if(cur->right) q.push(cur->right);
}
}

curDepth++;
}

}
};
``````

Code (Python)

``````class Solution(object):
def zigzagLevelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
if not root: return []
res, temp, q, flag=[], [], [root], 1
while q:
for i in xrange(len(q)):
node=q.pop(0)
temp+=[node.val]
if node.left: q+=[node.left]
if node.right: q+=[node.right]
res+=[temp[::flag]] // flag decides what is the final order being added!
temp=[]
flag*=-1
return res
``````

Special Thanks fangkunjnsy for the reference